Problems on Distance Between Two Points



Solving the problems on distance between two points with the help of the formula, in the below examples use the formula to find distance between two points.



Worked-out problems on distance between two points:

1. Show that the points (3, 0), (6, 4) and (- 1, 3 ) are the vertices of a right-angled Isosceles triangle.

Solution:
Let the given points be A(3, 0), B (6, 4) and C (-1, 3). Then we have,

AB2 = (6 - 3)2 + (4 - 0)2 = 9 + 16 = 25;

BC2 = (-1 - 6)2 + (3 - 4 )2 = 49 + 1= 50

and CA2 = (3 + 1)2 + (0 - 3)2 = 16 + 9= 25.

From the above results we get,

AB2 = CA2 i.e., AB = CA,

which proves that the triangle ABC is isosceles.

Again, AB2 + AC2 = 25 + 25 = 50 = BC2

which shows that the triangle ABC is right-angled.

Therefore, the triangle formed by joining the given points is a right-angled isosceles triangle. Proved.


2. If the three points (a, b), (a + k cos α, b + k sin α) and (a + k cos β, b + k sin β) are the vertices of an equilateral triangle, then which of the following is true and why ?

(i) | α - β| = π/4

(ii) |α - β| = π/2

(iii) |α - β| = π/6

(iv) |α - β| = π/3

Solution:
Let the vertices of the triangle be A (a, b), B (a + k cos α, b + k sin α) and C (a + k cos β, b + k sin β).

Now, AB2 = (a + k cos α - a)2 + (b + k sin α - b)2

= k2 cos2 α + k2 sin2 α = k2;

Similarly, CA2 = k2 and

BC2 = (a + k cos β - a - k cos α)2 + (b + k sin β - b - k sin α)2

= k2 (cos2 β + cos2 α - 2 cos α cos β + sin2 β + sin2 α - 2 sin α sin β)

= k2 [cos2 β + sin2 β + cos2 α + sin2 α - 2(cos α cos β + sin α sin β)]

= k2 [1 + 1 - 2 cos (α - β)]

= 2k2 [1 - cos (α - β)]

Since ABC is an equilateral triangle, hence

AB2 = BC2

or, k2 = 2k2 [1 - cos (α - β)]

or, 1/2 = 1 - cos(α - β) [since, k # 0]

or, cos (α - β) = 1/2 = cos π/3

Therefore, |α - β| = π/3 .

There for, condition (iv) is true.


3. Find the point on the y-axis which is equidistant from the points (2, 3)and(-1, 2).

Solution:
Let P(0, y) be the required point on the y-axis and the given points are A (2, 3) and B(- 1, 2). By question,

PA = PB = PA2 = PB2

or, (2 - 0)2 + (3 - y)2 = (-1 - 0)2 + (2 – y)2

or, 4 + 9 + y2 - 6y = 1 + 4 + y2 - 4y

or, - 6y + 4y = 1 - 9 or, - 2y = -8

or, y = 4.

Therefore, the required point on the y-axis is (0, 4).


4. Find the circum-centre and circum-radius of the triangle whose vertices are (3, 4), (3, - 6) and (- 1, 2).

Solution:
Let A(3, 4), B (3, - 6), C (- 1, 2) be the vertices of the triangle and P(x, y ) the required circum-centre and r the circum-radius. Then, we must have,

r2 = PA2 = (x - 3)2 + (y - 4)2 ……………………..(1)

r2 = PB2 = (x - 3)2 + (y + 6)2 ……………………….(2)

and r2 = PC2 = (x + 1)2 + (y - 2)2 ……………………….(3)

From (1) and (2) we get,

(x - 3)2 + (y - 4)2 = (x - 3)2 + (y + 6)2

Or, y2 - 8y + 16 = y2 + 12y + 36

or, - 20y = 20 or, y = - 1

Again, from (2) and (3) we get,

(x - 3)2 + (y + 6)2 = (x + 1 )2 + (y - 2)2

or, x2 - 6x + 9 + 25 = x2 + 2x + 1 + 9 [putting y = - 1]

or, - 8x = - 24

or, x = 3

Finally, putting x = 3 and y = - 1 in (1) we get,

r2 = 02 + (-1 - 4)2 = 25

Therefore, r = 5

Therefore, the co-ordinates of circum-centre are (3, - 1) and circum-radius = 5 units.


5. Show that the four points (2, 5), (5, 9), (9, 12) and (6, 8) when joined in order, form a rhombus.

Solution:
Let the given points be A(2, 5), B (5, 9), C (9, 12) and D(6, 8). Now,AB2 = (5 - 2)2 + (9 - 5)2 = 9 + 16 = 25

BC2 = (9 - 5)2 + (12 - 9)2 = 16 + 9 = 25

CD2 = (6 - 9)2 (8 - 12)2 = 9 + 16 = 25

DA2 = (2 - 6)2 + (5 - 8)2 = 16 + 9 = 25

AC2 = ( 9 - 2)2 + (12 - 5)2 = 49 + 49 = 98

and BD2 = (6 - 5)2 + (8 - 9)2 = 1 + 1 = 2

From the above result we see that

AB = BC = CD = DA and ACBD.

That is the four sides of the quadrilateral ABCD are equal but diagonals AC and BD are not equal. Therefore, the quadrilateral ABCD is a rhombus. Proved.

The above worked-out problems on distance between two points are explained step-by-step with the help of the formula.



Co-ordinate Geometry

  • What is Co-ordinate Geometry?
  • Rectangular Cartesian Co-ordinates
  • Polar Co-ordinates
  • Relation between Cartesian and Polar Co-Ordinates
  • Distance between Two given Points
  • Distance between Two Points in Polar Co-ordinates
  • Division of Line Segment: Internal & External
  • Area of the Triangle Formed by Three co-ordinate Points
  • Condition of Collinearity of Three Points
  • Medians of a Triangle are Concurrent
  • Apollonius' Theorem
  • Quadrilateral form a Parallelogram
  • Problems on Distance Between Two Points
  • Area of a Triangle Given 3 Points
  • Worksheet on Quadrants
  • Worksheet on Rectangular – Polar Conversion
  • Worksheet on Line-Segment Joining the Points
  • Worksheet on Distance Between Two Points
  • Worksheet on Distance Between the Polar Co-ordinates
  • Worksheet on Finding Mid-Point
  • Worksheet on Division of Line-Segment
  • Worksheet on Centroid of a Triangle
  • Worksheet on Area of Co-ordinate Triangle
  • Worksheet on Collinear Triangle
  • Worksheet on Area of Polygon
  • Worksheet on Cartesian Triangle

  • 11 and 12 Grade Math

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